$p:E\to B$ is fibration then $p_*:map(X,E)\to map(X,B)$ is fibration as well.

Question

$p:E\to B$ is fibration then for $X$ being compactly generated weakly Hausdorff space $p_*:map(X,E)\to map(X,B)$ is fibration as well.

We'd like to show that for any $Y$ and continuous $f$ and homotopy $H$ exists lift $\tilde H$ s.t. following diagram is commutative: $$ \begin{matrix} Y & \xrightarrow{f} & map(X,E) \\ \left\downarrow{i_0}\vphantom{\int}\right. & \nearrow{\tilde H}\vphantom{\int}& \left\downarrow{p_*}\vphantom{\int}\right.\\ Y\times I& \xrightarrow{H} & map(X,B) \end{matrix} $$ My attempt was to construct $\tilde H$ using fact that $p$ is fibration, namely for every element $y \in Y$ (which can be done because $Y$ is set):

$$ \begin{matrix} X & \xrightarrow{f(y)} & E \\ \left\downarrow{i_0}\vphantom{\int}\right. & \nearrow{\tilde H(y,\square)}\vphantom{\int}& \left\downarrow{p}\vphantom{\int}\right.\\ X\times I& \xrightarrow{H(y,\square)} & B \end{matrix} $$

where $H(y,\square): (x,t)\mapsto H(y,t)(x)$. Then homotopy given by formula $\tilde H: (y,t) \mapsto \tilde H(y,t)$ gives us demanded lift in first diagram. Is it correct proof? Where did I use (indirectly) assumption about $X$ being CGWH space or one can get rid of it? My suspicion suggest that this assumption could have used if we had consider all those diagrams in $\mathcal{Top}$ category not $\mathcal{Set}$ (I don't know in which category this problem should hold but the assumption suggests that it may be $\mathcal{Top}$). I'm not familiar with CGWH spaces so I'd be glad for stressing where this assumption plays a role.

Answer 1

Expanding on the comment of Adeel, you have to exploit the following easy fact about orthogonal classes of arrows in categories:

Let $F\dashv G$ be two adjoint functors between categories $\mathcal{C}\leftrightarrows\mathcal{D}$; then $Ff\perp g$ in the category $\mathcal D$ (i.e., $Ff$ has the LLP with respect to $g$ in $\cal D$) if and only if $f\perp Gg$ in the category $\cal C$.

Now apply this to $X\times(-)\dashv Map(X,-)$; the exercise is not finished (it remains showing that $X\times i$ is again an acyclic cofibration!) but I think it's the best hint to work conceptually.

Answer 2

Solution which I came up with today:

Let's recall the definition of a cocylinder $P(p)$ i.e. the pullback of such diagram: $$ \begin{matrix} P(p) & \xrightarrow{p'} & P(B) \\ \left\downarrow{p_0}\vphantom{\int}\right. & & \left\downarrow{p_0}\vphantom{\int}\right.\\ E& \xrightarrow{p} & B \end{matrix} $$ (where $p':(e,\omega)\mapsto p \circ \omega$). We may apply this for object $P(E)$ and morphisms $p_* :P(E) \to P(B)$, $p_0:P(E)\to E$ obtaining a unique arrow $\bar p :P(E) \to P(p)$. However we know that $p$ is fibration iff $\exists s:P(p) \to P(E)$ such that $ \bar p \circ s = id_{P(p)}$.

Let's map our diagram against functor $map(X, \square)$ obtaining:

$$ \begin{matrix} map(X,P(p)) & \xrightarrow{p'_*} & map(X,P(B)) \\ \left\downarrow{p_{0*}}\vphantom{\int}\right. & & \left\downarrow{p_{0*}}\vphantom{\int}\right.\\ map(X,E)& \xrightarrow{p_*} & map(X,B) \end{matrix} $$

There exists a unique arrow $ \bar p_* : P(map(X,E)) \to map(X,P(p))$ since $map(X,\square)$ preserves limits (hence pullbacks as well) (I'm cheating here a little because we know that there exists some unique arrow not necessarily $\bar p_*$ but it is isomorphic to $\bar p_*$ so let's stick to $\bar p_*$) but moreover in our CGWH category: $$map(X,P(B)) = map(X,map(I,B))\cong map(I \times X,B)\cong map(X \times I,B) \cong map(I ,map(X, B)) = P(map(X,B))$$ so $map(X,P(p))$ is cocylinder $P(p_*)$ and now we have only to show that there exists $s' : map(X,P(p)) \to P(map(X,E))$ such that $\bar p _* \circ s' = id_{P(p_*)}$. Surprisingly $s' = map(X,s)$ works like charm since $\bar p_* \circ map(X, s) = map(X,\bar p \circ s) = map(X,id_{P(p)}) = id_{map(X,P(p))} = id_{P(p_*)}$.

Answer 3

My attempt to prove categorical remarks of Adeel and tetrapharmakon from the comments was too long for the comment hence I'm posting it as an answer.

(Lemma) Let $F\dashv G$ be two adjoint functors between categories $\mathcal{C}\leftrightarrows\mathcal{D}$; then $Ff\perp g$ in the category $\mathcal D$ (i.e., $Ff$ has the LLP with respect to $g$ in $\cal D$) if and only if $f\perp Gg$ in the category $\cal C$. (where $f:C_1\to C_2$, $g:D_2 \to D_1$).

In other words there exists an arrow $FC_2 \to D_2$ making following diagram commutative: $$ \begin{matrix} FC_1 & \xrightarrow{u} & D_2 \\ \left\downarrow{Ff}\vphantom{\int}\right. & & \left\downarrow{g}\vphantom{\int}\right.\\ FC_2& \xrightarrow{v} & D_1 \end{matrix} $$

if and only if there exists an arrow $C_2 \to GD_2$ making following diagram commutative: $$ \begin{matrix} C_1 & \xrightarrow{u'} & GD_2 \\ \left\downarrow{f}\vphantom{\int}\right. & & \left\downarrow{Gg}\vphantom{\int}\right.\\ C_2& \xrightarrow{v'} & GD_1 \end{matrix} $$

But $F\dashv G$ means that following diagram is commutative:

$$ \begin{matrix} hom_\mathcal{D}(FC_2,D_2) & \xrightarrow{\simeq} & hom_\mathcal{C}(C_2,GD_2) \\ \left\downarrow{hom_\mathcal{D}(Ff,g)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(f,Gg)}\vphantom{\int}\right.\\ hom_\mathcal{D}(FC_1,D_1)& \xrightarrow{\simeq} & hom_\mathcal{C}(C_1,GD_1) \end{matrix} $$

Once again from adjunction (i.e. isomorphism between hom-sets) we know that to every $u' \in hom_\mathcal{D}(C_1,GD_2)$ corresponds uniquely some $u \in hom_\mathcal{D}(FC_1,D_2)$. Similarly let $v \in hom_\mathcal{C}(FC_2, D_1)$ be a morphism corresponding to $v' \in hom_\mathcal{C}(C_2,GD_1)$. Let's suppose that there exists $\eta:FC_2 \to D_2$ making the very first with diagram commutative for preceding choice of morphisms $u, v$ and $\phi:C_2 \to GD_2$ morphism corresponding to it by adjunction.

$$ \begin{matrix} hom_\mathcal{D}(FC_2,D_2) & \xrightarrow{\simeq} & hom_\mathcal{C}(C_2,GD_2) \\ \left\downarrow{hom_\mathcal{D}(Ff,id)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(f,id)}\vphantom{\int}\right.\\ hom_\mathcal{D}(FC_1,D_2)& \xrightarrow{\simeq} & hom_\mathcal{C}(C_1,GD_2) \\ \left\downarrow{hom_\mathcal{D}(id,g)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(id,Gg)}\vphantom{\int}\right.\\ hom_\mathcal{D}(FC_1,D_1)& \xrightarrow{\simeq} & hom_\mathcal{C}(C_1,GD_1) \end{matrix} $$ for setting above we have: $$ \begin{matrix} \eta & \xrightarrow{\simeq} & \phi \\ \left\downarrow{}\vphantom{\int}\right. & & \left\downarrow{}\vphantom{\int}\right.\\ \eta \circ Ff = u & \xrightarrow{\simeq} & \phi \circ f = u' \\ \left\downarrow{}\vphantom{\int}\right. & & \left\downarrow{}\vphantom{\int}\right.\\ g\circ u = g \circ \eta \circ Ff& \xrightarrow{\simeq} & Gg \circ \phi \circ f = Gg \circ u' \end{matrix} $$

Similarly: $$ \begin{matrix} hom_\mathcal{D}(FC_2,D_2) & \xrightarrow{\simeq} & hom_\mathcal{C}(C_2,GD_2) \\ \left\downarrow{hom_\mathcal{D}(id,Gg)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(id,g)}\vphantom{\int}\right.\\ hom_\mathcal{D}(FC_2,D_1)& \xrightarrow{\simeq} & hom_\mathcal{C}(C_2,GD_1) \\ \left\downarrow{hom_\mathcal{D}(Ff,id)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(f,id)}\vphantom{\int}\right.\\ hom_\mathcal{D}(FC_1,D_1)& \xrightarrow{\simeq} & hom_\mathcal{C}(C_1,GD_1) \end{matrix} $$ so: $$ \begin{matrix} \eta & \xrightarrow{\simeq} & \phi \\ \left\downarrow{}\vphantom{\int}\right. & & \left\downarrow{}\vphantom{\int}\right.\\ g \circ \eta = v & \xrightarrow{\simeq} & Gg \circ \phi = v' \\ \left\downarrow{}\vphantom{\int}\right. & & \left\downarrow{}\vphantom{\int}\right.\\ v \circ f = g \circ \eta \circ Ff& \xrightarrow{\simeq} & Gg \circ \phi \circ f = u' \circ f \end{matrix} $$ in the same way we can show that the existence of lifting in second diagram implies existence of lifting in the first and hence the thesis of lemma.